Riemann Integral - Is there a formula to calculate the following sum?

Question

I'm trying to calculate the integral of f(x)= x^2 between 1 and 2 by taking the limit of the Riemann sum. I couldn't find a way to factor all the constants out of the summation so I'm stuck with this: $$1/n^3\sum_{i=1}^{n}(n+i)^2$$ How do I go about it ?

Answer 1

Ok, so I tried working my way out using the distributive rule of summation: $$\sum_{i=1}^{n}(n+i)^2=n^2\sum_{i=1}^{n}1+2n\sum_{i=1}^{n}i+\sum_{i=1}^{n}i^2\\=n^3+\frac{2n^2(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}\\=\frac{14n^3+9n^2+n}{6}$$ And now I can multiply it by the constant term I factored out earlier, which is 1/n^3 : $$\frac{14n^3+9n^2+n}{6n^3}$$ All I have to do now is to take this sum to the limit: $$\int_{1}^{2}f(x)dx=\lim_{n\rightarrow \infty }\sum_{i=1}^{n}f(x_{i})\Delta x=\lim_{n\rightarrow \infty }\frac{14n^3+9n^2+n}{6n^3}=7/3$$

Answer 2

You can proceed as follows: \begin{align*} \frac{1}{n^3}\sum_{i=1}^n(n+i)^2 &= \frac{1}{n}\sum_{i=1}^n(1+i/n)^2\\ &\rightarrow\int_0^1(1+x)^2 dx\\ &=\int_1^2 x^2 dx. \end{align*}